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klemol [59]
3 years ago
7

1. Asphyxiation is a hazard posed by Compressed Natural Gas (CNG) vehicles and can be detected when you notice

Engineering
1 answer:
alexira [117]3 years ago
3 0

Smell of Mercaptan will allow us to  detect the Compressed natural gas. The smell is similar to the LPG used in our homes for cooking.

Explanation:

Compressed natural gas is heavier than air and tends to settle down near the ground. In case there is a leakage of CNG  the air near the ground surface will be displaced and get filled with CNG. This may result in aphyxiation i.e. lack of oxygen in enclosed places.

CNG is odorless and cannot be detected by smell. In order to make the detection possible, a small amount of mercaptan is added to CNG so that its smell is easily detected.

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4. What are these parts commonly called?
patriot [66]

These parts are commonly called carburetor emulsion tubes. These tubes maintain the air-fuel ratio at different speeds.

The carburetor is a device of the combustion engine power supply system that mixes fuel and air in order to facilitate internal combustion.

The carburetor emulsion tubes are tubes that maintain the air-fuel ratio at different velocities.

These tubes (carburetor emulsion tubes) are small brass cylinders where the metering needle slides into them.

Learn more about carburetors here:

brainly.com/question/4237015

7 0
2 years ago
Which statement about tensile stress is true? A. Forces that act perpendicular to the surface and pull an object apart exert a t
svp [43]

Answer:

A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object.

Explanation:

Tensile stress is referred as a deforming force, in which force acts perpendicular to the surface and pull an object apart, attempting to elongate it.

The tensile stress is a type of normal stress, in which a perpendicular force creates the stress to an object’s surface.

Hence, the correct option is "A."

3 0
3 years ago
For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio
diamong [38]

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

x^2 / Dt = constant

where D is constant

then

x^2 / t = constant

x^2_1 / t₁ = x^2_2 / t₂

x^2_1 t₂ = t₁x^2_2

t₂ = t₁x^2_2 / x^2_1

t₂ = (x^2_2 / x^2_1)t₁

t₂ = ( x_2 / x_1 )^2 × t₁

so we substitute

t₂ = ( 0.0049  / 0.0018  )^2 × 11.3 hrs

t₂ = 7.41 × 11.3 hrs

t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

8 0
2 years ago
A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The n
Vikki [24]

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

6 0
2 years ago
Find E[x] when x is sum of two fair dice?
Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

36

1

​

P(X=3)=P(1,2)+P(2,1)=

36

2

​

=

18

1

​

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

36

3

​

=

12

1

​

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

36

4

​

=

9

1

​

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

36

5

​

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

36

6

​

=

6

1

​

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

36

5

​

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

36

4

​

=

9

1

​

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

36

3

​

=

12

1

​

P(X=11)=P(5,6)+P(6,5)=

36

2

​

=

18

1

​

P(X=12)=P(6,6)=

36

1

​

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

​

⋅P(X

i

​

)

=2×

36

1

​

+3×

18

1

​

+4×

12

1

​

+5×

9

1

​

+6×

36

5

​

+7×

6

1

​

+8×

36

5

​

+9×

9

1

​

+10×

12

1

​

+11×

18

1

​

+12×

36

1

​

=

18

1

​

+

6

1

​

+

3

1

​

+

9

5

​

+

6

5

​

+

6

7

​

+

9

10

​

+1+

6

5

​

+

18

11

​

+

3

1

​

=7

E(X

2

)=∑X

i

2

​

⋅P(X

i

​

)

=4×

36

1

​

+9×

18

1

​

+16×

12

1

​

+25×

9

1

​

+36×

36

5

​

+49×

6

1

​

+64×

36

5

​

+81×

9

1

​

+100×

12

1

​

+121×

18

1

​

+144×

36

1

​

=

9

1

​

+

2

1

​

+

3

4

​

+

9

25

​

+5+

6

49

​

+

9

80

​

+9+

3

25

​

+

18

121

​

+4

=

18

987

​

=

6

329

​

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

​

=

5.833

​

=2.415

4 0
2 years ago
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